Comprehensive Study Guide: Optimization and Curve Sketching

Profit vs. Price Variables: - Capital $P$ stands for total profit. - Small $p$ stands for price per unit. - The speaker emphasizes that these are distinct variables, though students may choose different letters (e.g., $J$ ) for profit as long as the distinction is maintained.
Substitution Process: - The formula linking quantity $q$ and price $p$ is $q = 120 - p$ . - To express the profit formula solely in terms of $p$ , substitute $(120 - p)$ everywhere $q$ appears.
Maximization Steps: - 1. Multiply out the simplified profit expression and collect like terms. - 2. Calculate the first derivative of the profit function with respect to price ( $P'(p)$ ). - 3. Set the first derivative to zero ( $P'(p) = 0$ ). - In the provided example, this yields a price $p = 70$ .
Calculating the Largest Profit: - Finding $p = 70$ only identifies the optimal price. - To find the actual maximum profit, you must plug $p = 70$ back into the original profit formula, not the derivative (which would necessarily equal zero).

Determining Maxima/Minima: - To verify $p = 70$ is a maximum, calculate the second derivative ( $P''(p)$ ). - In this case, $P''(p) = -2$ .
Interpretation of Constant Second Derivatives: - Even if the second derivative is a constant (e.g., $-2$ ) and contains no $p$ to plug the value into, it applies to all points on the curve. - Since $-2 < 0$ , the function is always concave down, meaning any critical point found is a maximum.
Mark Distribution for the Test: - Calculating the second derivative correctly (1 mark). - Stating that $P''(p) < 0$ (1 mark). - Providing a final conclusion that the point is a maximum (1 mark).

Substitution in Terms of $q$ : - One could alternatively rearrange $q = 120 - p$ to $p = 120 - q$ and substitute $p$ out of the profit formula. - Differentiating with respect to $q$ would yield an optimal quantity, which must then be plugged back into the price-quantity equation to find the final price.
Absolute vs. Local Maximums: - Technically, to find an absolute maximum, one must check the boundaries/endpoints of the domain. - In this context, the endpoints for $p$ are typically $p = 0$ (price is zero) and $p = 120$ (quantity is zero). - Example Calculation: - At $p = 0$ or $p = 120$ , profits may be negative (e.g., $-2400$ ). - Compare these values to the profit at $p = 70$ . The largest resulting value is the absolute maximum.

Function Analyzed: $f(x) = \frac{\ln(x^2)}{x^2}$ .
Determining the Domain: - Strategy: Start by assuming the domain is all real numbers, then subtract "errors." - Error 1: Denominator cannot be zero ( $x^2 \neq 0 \implies x \neq 0$ ). - Error 2: Logarithms cannot take zero or negative arguments ( $x^2 > 0$ ). - Because the $x$ is squared inside the natural log, negative numbers are allowed (they become positive before the log is applied). Only $x = 0$ is excluded. - Domain: All real numbers except zero ( $x \in \mathbb{R}, x \neq 0$ ).
X-Intercepts ( $y = 0$ ): - Set $0 = \frac{\ln(x^2)}{x^2}$ . - Multiply by $x^2$ to get $0 = \ln(x^2)$ . - Solve using the definition of a logarithm: $e^0 = x^2 \implies 1 = x^2$ . - $x = \pm 1$ .
Y-Intercepts ( $x = 0$ ): - To find y-intercepts, substitute $x = 0$ into the function. - Since $x = 0$ is undefined (division by zero and $\ln(0)$ ), there is no y-intercept.
Parity and Symmetry: - Test for even/odd by plugging in $(-x)$ . - $f(-x) = \frac{\ln((-x)^2)}{(-x)^2} = \frac{\ln(x^2)}{x^2} = f(x)$ . - Because $f(-x) = f(x)$ , the function is even. - An even function is symmetrical about the y-axis.

The First Derivative: The speaker references a derivative where the numerator is $2(1 - \ln(x^2))$ .
Solving for Stationary Points: - Set the numerator to zero: $1 - \ln(x^2) = 0 \implies \ln(x^2) = 1$ . - $e^1 = x^2 \implies x = \pm \sqrt{e}$ . - Decimal approximation: $\sqrt{e} \approx 1.65$ . - Corresponding y-values: Plug $\pm \sqrt{e}$ into $f(x)$ . - Numerator: $\ln((\sqrt{e})^2) = \ln(e) = 1$ . - Denominator: $(\sqrt{e})^2 = e$ . - Stationary points are at $(\sqrt{e}, \frac{1}{e})$ and $(-\sqrt{e}, \frac{1}{e})$ .
Singular Points vs. Asymptotes: - A singular point occurs where the derivative is undefined but the original function is defined (e.g., $y = \sqrt{|x|}$ at $x = 0$ ). - $x = 0$ makes the derivative of $f(x) = \frac{\ln(x^2)}{x^2}$ undefined, but since the original function is also undefined at $x = 0$ , it is an asymptote, not a singular point.

Critical values for the number line: $-\sqrt{e}$ , $0$ , and $\sqrt{e}$ .
Interval Test (using $f'(x) = \frac{2(1 - \ln(x^2))}{x^3}$ ): - $x < -\sqrt{e}$ : (e.g., $x = -3$ ): $\text{Numerator} (-) / \text{Denominator} (-) = (+)$ . Increasing. - $-\sqrt{e} < x < 0$ : (e.g., $x = -1$ ): $\text{Numerator} (+) / \text{Denominator} (-) = (-)$ . Decreasing. - $0 < x < \sqrt{e}$ : (e.g., $x = 1$ ): $\text{Numerator} (+) / \text{Denominator} (+) = (+)$ . Increasing. - $x > \sqrt{e}$ : (e.g., $x = 3$ ): $\text{Numerator} (-) / \text{Denominator} (+) = (-)$ . Decreasing.
Conclusion: - $(-\sqrt{e}, \frac{1}{e})$ is a local maximum. - $(\sqrt{e}, \frac{1}{e})$ is a local maximum.

In the 2025 assessment, the second derivative and x-coordinates for inflection points were provided to students.
Inflection Points: Occur where $f''(x) = 0$ .
The speaker identifies the x-values for these points as $\pm \sqrt{e^{5/3}}$ (approx. $\pm 2.3$ ).
Number Line for $f''(x)$ : - Critical values: $-\sqrt{e^{5/3}}$ , $0$ , and $\sqrt{e^{5/3}}$ . - Even if a point is an asymptote (like $x = 0$ ), it must be included on the number line to check for changes in concavity.

Question: If zero is a singular point, would the graph have a sharp corner?
Response: If zero were a singular point, the graph would have a sharp corner or a vertical tangent. However, in this specific function, $x = 0$ is a vertical asymptote because the function itself is undefined there.
Question: Should we include the endpoint in the interval when stating where a function is increasing?
Response: Generally, functions are considered increasing on an interval if the derivative is $\ge 0$ . There is a technical distinction between "increasing" (includes $\ge 0$ ) and "strictly increasing" (only $> 0$ ), but usually, including the endpoint (e.g., $x \le -\sqrt{e}$ ) is acceptable and won't lose marks.